Given: $\{a,b\}\subset \mathbb R^+$, $P(x)=\frac{1}{x}+\frac{1}{x-a}+\frac{1}{x-b}=0.$ (correction: last term is $\frac{1}{x{\color{red}+}b})$

Show: all roots of $P(x)$ are real, one in the interval $(-b,0)$ and the other in the interval $(0,a)$.

Question asked in an entrance exam.

My attempt: developing $P(x)$, we get to $$x^2-\frac{2}{3}(a+b)x+\frac{ab}{3}=0$$ with discriminant defined by

$$\vartriangle=\frac{4}{9}(a+b)^2-\frac{4}{3}ab=\frac{4}{9}((a+b)^2-3ab)=\frac{4}{9}(a^2+b^2-ab)$$ As it is easy to show that $a^2+b^2-ab\ge 0$ it follows that $\vartriangle\ge 0$. This solves the first part of the question.

I tried, unsuccessfully, to solve the second part, which asked to show that the roots are within two specific intervals. Hints and full answers are welcome.

Edit 1. The term $\frac{1}{x-b}$ is, most likely, $\frac{1}{x+b}$. See comments below for an explanation.

  • $\begingroup$ I think you should check the original problem statement. As you copied it here, it is not correct. Possible the term $1/(x-b)$ should have been $1/(x+b)$. $\endgroup$ – hardmath Oct 21 '17 at 2:47
  • $\begingroup$ @hardmath, you are right, the statement I presented was copied from a book. I searched the original exam (IME exam 1976/77) and by chance I found the original question with an answer. The statement in the material is identical to the one I transcribed here, but there is a note stating that the solution assumes $1/(x+b)$ instead of $1/(x-b)$. As this question was not considered invalid, perhaps the original text had $1/(x-b)$ and the correction was given in the classroom during the exam (a guess). I will, later on, transcribe the solution presented there which uses calculus arguments. $\endgroup$ – bluemaster Oct 21 '17 at 12:56

I think the problem is not true. When a is smaller than b, consider the quadratic function you get , let x equal to 0, a and b, you will know that one root lies between 0 and a, another root lies between a and b. When a is bigger than b, it is similar.

  • $\begingroup$ Underlying your explanation that "the problem is not true" is the symmetry between the roles of $a$ and $b$. As you say, "When $a$ is bigger than $b$, it is similar" to the case when $a$ is smaller than $b$. Yet the problem as stated draws a conclusion (one root in $(-b,0)$ and the other in $(0,a)$) in which the roles of $a$ and $b$ are not interchangeable. $\endgroup$ – hardmath Oct 21 '17 at 2:28
  • $\begingroup$ I don't use the symmetry between a and b. I discussed about which one is bigger in order to show where are the two roots. Then I get their locations. However, my conclusion is different from that in the provlem. So I think the problem has some mistakes. $\endgroup$ – Jiongjie Wang Oct 21 '17 at 2:41
  • $\begingroup$ I realize that you don't use the symmetry between $a$ and $b$, or more precisely, between the roles they play in the problem setup (both are positive numbers, and the terms $1/(x-a)$ plus $1/(x-b)$ commute if $a,b$ are exchanged). However I thought I should point out the symmetry to show that yes, the problem is wrong because it proposes that even though the roles of $a,b$ are interchangeable, a conclusion is proposed that has $a,b$ playing distinguised roles. $\endgroup$ – hardmath Oct 21 '17 at 2:44
  • $\begingroup$ Yes, I agree with you. Your thoughts are helpful! $\endgroup$ – Jiongjie Wang Oct 21 '17 at 2:50
  • $\begingroup$ JiongjieWang and @hardmath you are right... see comments below problem statement. $\endgroup$ – bluemaster Oct 21 '17 at 13:07

I am using calculus. Please tell me if it is not allowed in the exam, then I can post an alternate solution.

It is actually very simple with calculus : $P(x)$ is a continuous function on $\mathbb R \backslash\{0,a,b\}$. Without loss of generality, let us assume that $0 < a < b$. Then, in the interval $(0,a)$, note that as $x \to 0^+$, $P(x)$ eventually becomes positive, and as $x \to a^-$, $P(x)$ eventually becomes negative, and therefore, by the intermediate value theorem, $P(x)$ has a root in $(0,a)$. A similar logic shows that it has a root in $(a,b)$.

Finally, we note that for $x < 0$, all three quantities $\frac{1}{x},\frac{1}{x-a},\frac{1}{x-b}$ are strictly negative, hence their sum is strictly negative. Similarly, for $x>b$, the sum is strictly positive. Hence, no roots lie in these areas, and therefore there are precisely two roots of this equation, both real, by your working on the discriminant.

  • $\begingroup$ No. This exam considers algebra precalculus. But your argument is nice and is worth keeping here. $\endgroup$ – bluemaster Oct 21 '17 at 1:47
  • $\begingroup$ Yes, but somewhere I find ti difficult to recollect, how we used to prove existence of roots in precalculus algebra. I mean, if you say, show that a root of some equation lies between some two given points, then what else, other than the IVT, could I use to show just the existence of the root? For one, it may not be easy to compute the root, for example say something like $x = \cos x$ has a root in $[-2,2]$. $\endgroup$ – астон вілла олоф мэллбэрг Oct 21 '17 at 1:51
  • $\begingroup$ See comments below the problem statement. The "official" solution uses calculus arguments (I will transcribe it later on) but states that the last term is not $\frac{1}{x-b}$ but $\frac{1}{x+b}$. This exam is known for being very tough... and often asks material beyond the scope of the official syllabus. $\endgroup$ – bluemaster Oct 21 '17 at 13:05

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